The 7-digit number 93A562B is divisible by 9, What is the minimum value of (A + B)?

Correct Answer: Option 2 (which gives the minimum value of A + B as 2)

Step-by-Step Explanation:

1. Divisibility Rule of 9:
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

2. Sum of the Digits:
The given 7-digit number is 93A562B.
Let us find the sum of all its digits:

$\text{Sum of digits}=9+3+A+5+6+2+B$

$\text{Sum of digits}=25+A+B$

3. Finding the Minimum Value of (A + B):
For the number to be divisible by 9, the sum $(25+A+B)$ must be a multiple of 9 (e.g., 27, 36, 45, etc.).

Since A and B are single digits (ranging from 0 to 9), $A+B$ must be a non-negative integer.

The smallest multiple of 9 that is greater than or equal to 25 is 27.

Set the sum equal to 27:

$25+A+B=27$

Subtract 25 from both sides:

$A+B=27-25$

$A+B=2$

Therefore, the minimum value of (A + B) is 2.